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Michele_Laino

  • one year ago

Tutorial on the Quaternions of Hamilton

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  1. Michele_Laino
    • one year ago
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  2. Astrophysics
    • one year ago
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    Thank you Michele, I will give it a read! :)

  3. Michele_Laino
    • one year ago
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    thanks!! :) @Astrophysics

  4. Oleg3321
    • one year ago
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    is this some sort of new language??? lol

  5. Michele_Laino
    • one year ago
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    no, the quaternions were introduced by W. R. Hamilton in the far 1833 :) lol

  6. Oleg3321
    • one year ago
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    lol looks very complicated

  7. Michele_Laino
    • one year ago
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    no, they are not complicated, please study my definitions first @oleg3321

  8. Astrophysics
    • one year ago
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    Very easy to understand, thanks again!

  9. Michele_Laino
    • one year ago
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    :) @Astrophysics

  10. welshfella
    • one year ago
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    Interesting article. Well explained. Hamilton was Ireland's most famous mathematician . Unfortunately his life was cut short by alcoholism.

  11. Michele_Laino
    • one year ago
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    Thanks! @welshfella

  12. DanJS
    • one year ago
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    I remember my calc 3 teacher mentioning those to look at if you got bored.. thanks @Michele_Laino

  13. Michele_Laino
    • one year ago
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    thanks! :) @DanJS

  14. arindameducationusc
    • one year ago
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    Thank you @Michele_Laino . I will definitely give it a read

  15. Michele_Laino
    • one year ago
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    Thanks! :) @arindameducationusc

  16. anonymous
    • one year ago
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    quaternions aren't a field, they are a division ring

  17. Michele_Laino
    • one year ago
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    Quaternions are a field @oldrin.bataku

  18. anonymous
    • one year ago
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    also you should introduce those other basis matrices \(\mathbf i,\mathbf k\) more naturally -- something like this: consider for \(z=a+bi\) we have: $$\begin{align*}a+bi\mapsto \begin{pmatrix}a+bi&0\\0&a-bi\end{pmatrix}&=\begin{pmatrix}a&0\\0&a\end{pmatrix}+\begin{pmatrix}bi&0\\0&-bi\end{pmatrix}\\&=a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}i&0\\0&-i\end{pmatrix}\\&=a\cdot\mathbf1+b\cdot\mathbf i\end{align*}$$ and then we have that \(\mathbf i\cdot\mathbf j=\mathbf k\)

  19. anonymous
    • one year ago
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    https://en.wikipedia.org/wiki/Quaternion#Noncommutativity_of_multiplication >The fact that quaternion multiplication is not commutative makes the quaternions an often-cited example of a strictly skew field.

  20. Michele_Laino
    • one year ago
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    I wrote that formula: \[{\mathbf{ij}} = - {\mathbf{ji}} = {\mathbf{k}}\]

  21. anonymous
    • one year ago
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    https://en.wikipedia.org/wiki/Division_ring >In abstract algebra, a division ring, also called a skew field, is a ring in which division is possible. >Division rings differ from fields only in that their multiplication is not required to be commutative. ... Historically, division rings were sometimes referred to as fields, while fields were called “commutative fields”.

  22. anonymous
    • one year ago
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    yes, I know you did, but you introduced \(\mathbf k\) as a matrix first without any motivation. the reason for its existence is that we have four linearly independent matrices \(\mathbf {1,i,j,ij}\) since the multiplication is bilinear, and then we just call \(\mathbf {ij}=\mathbf k\)

  23. Michele_Laino
    • one year ago
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    I never worked with non-commutative algebra

  24. Michele_Laino
    • one year ago
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    I meant I never cited non-commutative algebra

  25. anonymous
    • one year ago
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    ? fields have commutative multiplication by definition, whereas the quaternions \(\mathbb H\) are a non-commutative algebra over \(\mathbb R\) and a division ring, not a field

  26. Michele_Laino
    • one year ago
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    when I wrote a matrix in M2(C) it is supposed that the multiplication is non-commutative since as you know, the multiplication betwen matrices is non-commutative

  27. anonymous
    • one year ago
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    lol, yes, clearly, and these matrices do not commute, so they cannot form a field. the quaternions are not a field.

  28. zzr0ck3r
    • one year ago
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    :) right

  29. zzr0ck3r
    • one year ago
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    For real man, tuts that are wrong are even worse. This is all simple definition stuff you are getting wrong...Please remove.

  30. Michele_Laino
    • one year ago
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    More precisely, I have used the term "field", since it is commonly used in mathematics literature, a more appropriate term can be "sfield" which indicate a non-commutative field. I have supposed that, being the non-commutative characteristic evident, there was no need to use the term "sfield"

  31. anonymous
    • one year ago
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    field is not commonly used in mathematics literature for a skew field, this isn't the 1800s anymore

  32. anonymous
    • one year ago
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    it's called a division ring, that is the appropriate, correct, and commonly used term for algebraic structures like this

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